Problem: Solve for $e$. $\dfrac43=-6e-\dfrac53 $
Let's add and then divide to get $e$ by itself. $\begin{aligned} \dfrac43&=-6e-\dfrac53 \\ \\ \dfrac43{+\dfrac53} &=-6e-\dfrac53 {+\dfrac53}~~~~~~~~{\text{add }\dfrac53} \text{ to each side}\\ \\ \dfrac43{+\dfrac53}&=-6e-\cancel{ \dfrac53} {{+}\cancel{{\dfrac53}}}\\ \\ \dfrac43{+\dfrac53}&=-6e \end{aligned}$ $\begin{aligned}3&= -6e \\ \\ \dfrac{3}{{-6}} &= \dfrac{-6e}{{-6}} ~~~~~~~\text{divide each side by } {-6} \text{ to get } e \text{ by itself }\\ \\ \dfrac{3}{{-6}} &= \dfrac{\cancel{-6}e}{\cancel{{-6}}} \\ \\ \dfrac{3}{{-6}}&= e \end{aligned}$ The answer: $e={-\dfrac12}$ Let's check to make sure. $\begin{aligned} \dfrac43&=-6e-\dfrac53\\\\ \dfrac43&\stackrel{?}{=} -6\left({-\dfrac12}\right)-\dfrac53 \\\\ \dfrac43&\stackrel{?}{=} 3-\dfrac53 \\\\ \dfrac43 &= \dfrac43 ~~~~~~~~~~\text{Yes!} \end{aligned}$